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CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts f (x) g’(x) dx = f (x) g(x) - g(x) f’(x) dx . Substitution Rule that is easy to remember Let u = f (x) and v = g(x). Then the differentials are du = f’(x) dx and dv = g’(x) dx and the formula is: u dv = u v - v du .

CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts f (x) g’(x) dx = f (x) g(x) - g(x) f’(x) dx. Substitution Rule that

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Page 1: CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts  f (x) g’(x) dx = f (x) g(x) -  g(x) f’(x) dx. Substitution Rule that

CHAPTER 2 2.4 Continuity

Integration by Parts

The formula for integration by parts

f (x) g’(x) dx = f (x) g(x) - g(x) f’(x) dx .

Substitution Rule that is easy to remember

Let u = f (x) and v = g(x). Then the differentials are du = f’(x) dx and dv = g’(x) dx and the formula is:

u dv = u v - v du .

Page 2: CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts  f (x) g’(x) dx = f (x) g(x) -  g(x) f’(x) dx. Substitution Rule that

Example: Find x cos x dx.

Example: Evaluate t2 e

t dt.

a

b f (x) g’(x) dx = [f (x) g(x)]a

b - a

b g(x)

f ’(x) dx.

Example: Evaluate 01 tan

-1x dx.

Example: Evaluate x ln x dx.

Example: Evaluate (2x + 3) ex

dx.

Page 3: CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts  f (x) g’(x) dx = f (x) g(x) -  g(x) f’(x) dx. Substitution Rule that

a

b f (x) g’(x) dx = [f (x) g(x)]a

b - a

b g(x)

f ’(x) dx.

Example: Evaluate 01 tan

-1x dx.

Example: Evaluate 1e x

ln x dx.

Example: Evaluate 01(2x + 3)

ex dx.

Page 4: CHAPTER 2 2.4 Continuity Integration by Parts The formula for integration by parts  f (x) g’(x) dx = f (x) g(x) -  g(x) f’(x) dx. Substitution Rule that

CHAPTER 2 2.4 ContinuityIntegration Using Technology

and Tables

Example: Use the Table of Integrals to find:

2. [(4 - 3x2 )0.5 / x ] dx.

3. e sin x sin 2x dx.

1. x2 cos 3x dx.